Mathematics - Determinant Question with Solution | TestHub

$\left|\begin{array}{lll}1& 2\mathrm{a}& \mathrm{a}\\ 1& 3\mathrm{b}& \mathrm{b}\\ 1& 4\mathrm{c}& \mathrm{c}\end{array}\right|=0{\mathrm{C}}_2\to {\mathrm{C}}_2-2{\mathrm{C}}_3$ $\left|\begin{array}{ccc}1& 0& \mathrm{a}\\ 1& \mathrm{b}& \mathrm{b}\\ 1& 2\mathrm{c}& \mathrm{c}\end{array}\right|=0{\mathrm{R}}_3\to {\mathrm{R}}_3-{\mathrm{R}}_2,{\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_1$ $\left|\begin{array}{ccc}1& 0& a\\ 0& b& b-a\\ 0& 2c-b& c-b\end{array}\right|=0$ $b(c-b)-(b-a)(2c-b)=0$ On simplification, $$\frac{2}{\mathrm{b}}=\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{c}}$$ $\therefore \mathrm{a},\mathrm{b},\mathrm{c}$ are in Harmonic Progression.